\documentclass[12pt, a4paper, oneside]{ctexart}
\usepackage{amsmath, amsthm, amssymb, bm, color, framed, graphicx, hyperref, mathrsfs}

\title{\textbf{一个微分中值定理的妙用}}
\author{张浩然}
\date{\today}
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\begin{document}

\maketitle



\begin{problem}

设 $a_i, b_i \geq M > 0$，且 $|a_i - b_i| \leq \varepsilon$，
其中 $i = 1, 2, \ldots, n$.证明：
$$
\left|\prod_{i=1}^{n}\dfrac{1}{a_i} -\prod_{i=1}^{n}\dfrac{1}{b_i} \right|\leqslant \dfrac{n\varepsilon}{M^{n+1}}
$$

\end{problem}

\begin{solution}
  \par
  我们考虑一个多元函数，$$
f(x)=\prod_{i=1}^{n}\dfrac{1}{x_i},\quad 
$$
\par
其中，
$$
x=(x_1,x_2,\cdots,x_n)^t, x_i\geqslant M, 1\leqslant i\leqslant n
  $$
\par
考虑多元函数微分中值定理：
$$
f(a)-f(b)=\nabla f(\xi)\cdot(a-b)
$$
\par
展开为:
$$
\prod_{i=1}^{n}\dfrac{1}{a_i} -\prod_{i=1}^{n}\dfrac{1}{b_i}=\sum_{i=1}^{n}-\dfrac{a_i-b_i}{\xi_i^2}\cdot\prod_{1\leqslant k\leqslant n, k\neq i}\dfrac{1}{\xi_k}
$$
\par
其中，$$
\xi_i=b_i+\theta(a_i-b_i), \theta \in(0,1), \xi_i\geqslant M, |a_i-b_i|\leqslant \varepsilon
$$
\par
于是：
$$
\begin{aligned}
\left|\prod_{i=1}^{n}\dfrac{1}{a_i} -\prod_{i=1}^{n}\dfrac{1}{b_i}\right|
&=\sum_{i=1}^{n}\dfrac{|a_i-b_i|}{\xi_i^2}\cdot\prod_{1\leqslant k\leqslant n, k\neq i}\dfrac{1}{|\xi_k|}\\
&\leqslant \sum_{i=1}^{n} \dfrac{\varepsilon}{M^{n+1}}\\
&=\dfrac{n\varepsilon}{M^{n+1}}
\end{aligned}
$$



\end{solution}
\end{document}